Quantum Mechanics - Probability

The wave function for a particle constrained to move in one dimension is shown in the graph (Ψ = 0 for x ≤  0 and x ≥ 5). What is the probability that the particle would be found between x = 2 and x = 4?

A. 17/64
B. 25/64
C. 5/8
D. v(5/8)
E. 13/16

(GR9677 #17)

Solution:

Probability, P ∼ ψ²

Probability to find the particle between x = 2 and x = 4 (unnormalized probability):
ψ² = 2²  + 3² = 13

Total probability (normalized probability):
ψ² = 1²  +  1²  + 2²  + 3²  +  1² = 16

P = unnormalized probability / normalized probability = 13/16

Answer: E

Quantum Mechanics - Potential Wall

Consider a potential of the form

V(x) = 0, x ≤ a
V(x) = V_0, a < x < b
V(x) = 0, x ≥ b

As shown in the figure above. Which of the following wave functions is possible for a particle incident from the left with energy E < V₀.

(GR9677 #18)

Solution:

A. Classic not QM potential → FALSE
B. No decrease in amplitude → FALSE
C. Decay exponentially inside the wall, decrease amplitude (fits E < V₀) →  TRUE

D. QM Oscillator harmonics → FALSE
E. Cosine wave function not QM potential → FALSE

Answer: C

Quantum Mechanics - Planck Length

The characteristic distance at which quantum gravitational effects are significant, the Planck length can be determined from a suitable combination of the physical constants G, ħ, and c. Which of the following correctly gives the Planck length?

A. Għc
B. Għ²c³
C. G²ħc
D. G^½ħ²c
E. (Għ/c³)^½

(GR9677 #29)

Solution:

Check units:

Let's l = Planck length
 l = G^x ħ^y c^z

G unit = m³ kg⁻¹ s⁻²
ħ unit = Js = Nms = kg m² s⁻¹
c unit = m s⁻¹
l unit = m

m = (m³ kg⁻¹ s⁻²)^x  (kg m² s⁻¹)^y (m s⁻¹)^z

For m:
1 = 3x + 2y + z

For kg:
0 = − x + y
x = y

For s:
0 = − 2x − y − z
0 = − 3x − z

z = − 3x

1 = 3x + 2y + z 

1 = 3x + 2x − 3x = 2x
x = y = 1/2
z = − 3x = − 3/2

l = G^x ħ^y c^z 
= G^1/2 ħ^1/2 c^−3/2
= (Għ/c³)^½

Answer: E

Quantum Mechanics - Spherical Harmonics

A diatomic molecules is initially in the state Ψ(Θ, Φ) = (5Y₁¹ + 3Y₅¹ + 2Y₅⁻¹) / (38)^½ where Y_l^m is a spherical harmonics. If measurements are made of the total angular momentum quantum number l and of azimuthal angular momentum quantum number m, what is the probability of obtaining the results l = 5?

A. 36/1444
B. 9/38
C. 13/38
D. 5/(38)^½
E. 34/38

(GR9677 #33)

Solution:

l = 5 → Y₅¹ and Y₅⁻¹

Probability, P = ∑_i |c_i|² 

P = |3/√38|² + |2/√38|² =  9/38 + 4/38 =  13/38

Answer: C

Quantum Mechanics - Fermion

The wave function for identical fermions is antisymmetric under particle interchange. Which of the following is a consequence of this property?

A. Pauli exclusion principle
B. Bohr correspondence principle
C. Heisenberg uncertainty principle
D. Bose-Einstein condensation
E. Fermi's golden rule

(GR9677 #35)

Solution:

Pauli Exclusion Principle: no two electrons can have exactly the same quantum number.

m_s1 =  ½  and m_s2 =  −½
Total spin quantum number, s = ½ + (−½) = 0
Multiplicity, 2 · 0 + 1 = 1 → Singlet state (antisymmetric)

Answer: A

Notes:

Singlet: 2s + 1 = 1, s = 0
Singlet state is anti-symmetric: ψ(1,2) = −ψ(2,1)
Obeys Fermi-Dirac statistics → fermion

Triplet: 2s + 1 = 3, s = 1
Triplet state is symmetric: ψ(1,2) = ψ(2,1)
Obeys Bose-Einstein statistics → bosons

Quantum Mechanics - Ionization Potential

Which of the following atoms has the lowest ionization potential?

A. ²_₄He
B. ⁷₁₄N
C. ⁸₁₆O
D. ¹⁸₄₀Ar
E. ⁵⁵₁₃₃Cs

(GR9677 #39)

Solution:

Ionization potential: the energy required to to remove the valence (outermost) electron from an atom

It is minimum for atoms (alkali metals) which have weakly bound electrons and higher for the noble gases which have closed shells.

²_₄He and Argon ⁸₄₀Ar are noble gases which have a closed-shell valence electron configuration.

⁷₁₄N → 1s² 2s² 2p⁶ 3s² 3p² (2 outermost electrons, almost a full orbital)

⁸₁₆O → 1s² 2s² 2p⁶⁴ 3s² 3p² (4 outermost electrons, almost a full orbital)

⁵⁵₁₃₃Cs has the highest atomic number and hence a lot of electrons. The valence electron is far from the nucleus (weakly bound), thus it needs the lowest ionization potential to remove it.

Answer: E

Quantum Mechanics - Spectroscopic Notation

A 3p electron is found in the ³P_3/2 energy level of a hydrogen atom. Which of the following is true about the electron in this state?

A. It is allowed to make an electric dipole transition to the ²S_1/2 level
B. It is allowed to make an electric dipole transition to the ²P_1/2 level
C. It has quantum numbers l = 3, j = 3/2, s = 1/2
D. It has quantum numbers n = 3, j = l, s = 3/2
E. It has exactly the same energy as it would in the ³D_3/2 level

(GR9677 #41)

Solution:

Spectroscopic notation: N^2s+1 L_j=l+s

For ³P_3/2  : 

P → l = 1
j = 3/2

→ (C) and (D) are FALSE

Selection rules (for electron transition):

1.	Principal quantum number	:	∆n = anything
2.	Orbital angular momentum	:	∆l = ±1
3.	Magnetic quantum number	:	∆ml = 0, ±1
4.	Spin	:	∆s = 0
5.	Total angular momentum	:	∆j = 0, ±1, but j = 0 ↛j = 0

(A) TRUE
Transition for ∆l = 0 is allowed.

(B) FALSE
Transition for ∆l = 0 is not allowed.

(E) FALSE
Electron can not move to different energy level (P to D) while maintaining the same energy.

Answer: A

Nuclear & Particle Physics - Photoelectric

Light of wavelength 500 nanometers is incident on sodium, with work function 2.28 electron volts. What is the maximum kinetic energy of the ejected photoelectrons?

A. 0.03 eV
B. 0.2 eV
C. 0.6 eV
D. 1.3 eV
E. 2.0 eV

(GR9677 #42)

Solution:

Einstein’s photoelectric equation:  |eV| = hv − W = ^hc/_λ − W

h = 4.1 × 10⁻¹⁵ eV.second
c = 3 × 10⁸ m/s
λ = 500 nm = 5 × 10⁻⁷ m
W = 2.28 eV

KE = [(4.1 × 10⁻¹⁵)(3 × 10⁸)/(5 × 10⁻⁷ )] − 2.28 
=  [(4.1 × 3 / 5)(10⁻¹⁵ × 10⁸ × 10⁷)] − 2.28
=  2.46 − 2.28 = 0.18 ≈  0.2 eV

Answer: B

Quantum Mechanics - Schrodinger Equation

The solution to the Schrödinger equation for a particle bound in a one-dimensional, infinitely deep potential well, indexed by quantum number n, indicates that in the middle of the well the probability density vanishes for

A. The ground state (n = 1) only
B. States of even n (n = 2, 4, ...)
C. States of odd n (n = 1, 3, ...)
D. All states (n = 1, 2, 3, ...)
E. All states except the ground state

(GR9677 #51)

Solution:

Wavefunctions for the first 5 states for a particle bound in a 1-D, infinitely deep potential well

Pic: ecee.colorado.edu

The even wave functions (n = 2, 4, ...) always have nodes in the middle.

→ probability density for states of even n vanishes.

Answer: B

Quantum Mechanics - Angular Momentum

At a given instant of time, a rigid rotator is in the state ψ(θ, ϕ) = √(¾π) sin θ sin ϕ, where θ is the polar angle relative to the z-axis and ϕ is the azimuthal angle. Measurement will find which of the following possible values of the z-component of the angular momentum L_z?

A. 0
B. ħ/2, −ħ/2
C. ħ, −ħ
D. 2ħ, −2ħ
E. ħ, 0, −ħ

(GR9677 #52)

Solution:

The eigenstates of L_z  is L_zY_l^m = mħY_l^m 

Possible values = eigen values of mħ.

L_z = −iħ ∂/∂ϕ → the information about m is contained in the sin ϕ term and it's proportional to e^imϕ:

sin ϕ  = (e^iϕ − e^−iϕ) / 2i

 → m = ±1

Thus, the eigenvalues are ±ħ

Answer: C

Quantum Mechanics - Laser

The approximate number of photons in a femtosecond (10⁻¹⁵ s) pulse of 600 nanometers wavelength light from a 10-kilowatt peak-power dye laser is

A. 10³
B. 10⁷
C. 10¹¹
D. 10¹⁵
E. 10¹⁸

(GR9677 #59)

Solution:

Number of photon, n = E_total / E_single

Energy of a single photon,  E_single = hc/λ 

h = 6.63 × 10⁻³⁴ Js
c = 3 × 10⁸ m/s
λ  =  600 nanometers = 600 × 10⁻⁹ m

→ E_single = 3 × 10⁻¹⁹  J

From Power, P = W/t = E_total/t

→ Total energy, E_total =  Pt

P = 10 kW = 10⁴ W
t = 10⁻¹⁵ s

→ E_total = 10⁻¹¹ J

n = E_total / E_single = 10⁻¹¹  / (3 × 10⁻¹⁹)  = 0.3 × 10⁸ = 3 × 10⁷  J

Answer: B

Quantum Mechanics - Gaussian Wave Packet

A Gaussian wave packet travels through free space. Which of the following statement about the wave packet are correct for all such wave packets?

1. The average momentum of the wave packet is zero
2. The width of the wave packet increases with time, as t → ∞.
3. The amplitude of the wave packet remains constant with time.
4. The narrower the wave packet is in momentum space, the wider it is in coordinate space.

A. I and III only
B. II and IV only
C. I, II, and IV only
D. II,III, and IV only
E. I, II,III, and IV only

(GR9677 #76)

Solution:

The Gaussian wave packet satisfies the Heisenberg uncertainty principle: ΔxΔp ≥ ħ/2

→ Statement IV is TRUE.
→ Statement I is FALSE, Δp cannot be 0
→ Answer A, C, E are FALSE.

Answer D suggests that both II and III are true.
If II and III are true, the area of gaussian wave packet can go to ∞
This is not allowed by the Heisenberg uncertainty principle.
→ D is FALSE

Answer: B

Quantum Mechanics - Spin Angular Momentum

Two ions 1 and 2, at fixed separation, with spin angular momentum operators S₁ and S₂, have the interaction Hamiltonian H = −J S₁·S₂, where J > 0. The values of S₁² and S₂² are fixed at S₁(S₁ + 1) and S₂(S₂ + 1), respectively. Which of the following is the energy of the ground state of the system?

A. 0
B. –JS₁S₂
C. –J[S₁(S₁ + 1) – S₂(S₂ + 1)]
D. –(J/2)[(S₁ + S₂)(S₁ + S₂ + 1) – S₁(S₁+1) – S₂(S₂+1)]
E. –(J/2)[(S₁(S₁ + 1) + S₂(S₂ + 1))/(S₁ + S₂)(S₁ + S₂ + 1)]

(GR9677 #77)

Solution:

S₁² ψ = S₁(S₁ + 1) ψ
S₂² ψ = S₂(S₂ + 1) ψ
S_i² ψ = S_i(S_i + 1) ψ

H = −J S₁·S₂

Using general arithmetic equation: ab = ½ [(a + b)² − a² − b²]

H = −(J/2)[(S₁ + S₂)² − S₁² − S₂²]

Since S_i² ψ = S_i(S_i + 1) ψ
For (S₁ + S₂)² → replace S_i with S₁ + S₂
→ (S₁ + S₂)² ψ = (S₁ + S₂)(S₁ + S₂ + 1) ψ

H = −(J/2)[(S₁ + S₂)(S₁ + S₂ + 1) − S₁(S₁ + 1) − S₂(S₂ + 1)]

Answer: D 

Quantum Mechanics - Momentum Operator

The wave function of a particle is e^i(kx−ωt) where x is distance, t is time, and k and are ω positive real numbers. The x–component of the momentum of the particle is 

A. 0
B. ħω
C. ħk
D. ħω/c
E. ħk/ω 

(GR9277 #01)

Solution: 

Momentum operator, p = −iħ∇

pψ = −iħ ∂ψ/∂x 

= −iħ ∂e^i(kx−ωt)/∂x 

= ħk e^i(kx−ωt) 

= ħkψ  

Answer: C 

Quantum Mechanics - Wave Function

If a freely moving electron is localized in space to within ∆x₀ of x₀, its wave function can be described by a wave packet

where f(k) is peaked around a central value k₀. Which of the following is most nearly the width of the peak in k?

A. 

B.

C.

D.

E.

(GR9277 #27)

Solution:

In quantum mechanics, the momentum p = ħk and position x wave functions are Fourier transform pairs and the relation between p and x representations forms the Heisenberg uncertainty relation:

∆x∆k ≥ 1 → ∆k ≥ 1/∆x

Or, since k and x are fourier variables, their localization would vary inversely.

Answer: B

Quantum Mechanics - Probability

A system is know to be in the normalized state described by the wave function



Where  are the spherical harmonics. The probability of finding the system in a state with azimuthal orbital quantum number m = 3 is

A. 0
B. 1/15
C. 1/6
D. 1/3
E. 13/15

(GR9277 #28)

Solution:

For the state  ,  probability is  

States with  m = 3 → and  with  c₁ = 5/√30 and c₂ = 1/√30

Therefore,



Answer: E

Quantum Mechanics, Particle in a box

An attractive, one-dimensional square well has depth V₀ as shown above. Which of the following best shows a possible wave function for a bound state?

(GR9277 #29)

Solution:

1-D square well with depth  V₀  → finite potential well.

Wave function of a particle in finite potential well should have these following properties:

• ψ →0 as x → ± ∞ or get further into regions where the classical particle cannot penetrate at all due to its inadequate energy
• ψ is always a decaying exponential function
• ψ must be continuous and differentiable

Therefore:

A. FALSE. ψ does not go to zero as x → ±∞

C. FALSE. ψ is not continuous

D and E are FALSE. ψ are not decaying exponential functions

Answer: B

Quantum Mechanics - Wave Function

The state of a quantum mechanical system is described by a wave function. Consider two physical observables that have discrete eigenvalues: observable A with eigenvalues {α}, and observable B with eigenvalues {β}. Under what circumstances can all wave functions be expanded in a set of basis states, each of which is a simultaneous eigenfunction of both A and B?

A. Only if the values {α} and {β} are nondegenerate
B. Only if A and B commute
C. Only if A commutes with the Hamiltonian of the system
D. Only if B commutes with the Hamiltonian of the system
E. Under all circumstances

(GR9277 #50)

Solution:

For two physical quantities to be simultaneously observable, their operator representations must commute, [A, B] = 0.

Answer: B


Proof:

Wave function, ψ = ∑_i c_i |v_i >

A |v_i > = α |v_i >

B |v_i > = β |v_i >

BA |v_i > = Bα |v_i > = α B |v_i > = αβ |v_i >

AB |v_i > = Aβ |v_i >= β A |v_i > = βα |v_i >

(BA − AB) |v_i > = (αβ − βα) |v_i > = 0

[A, B] = 0

Quantum Mechanics - Infinite Potential Well

Questions 51-53

A particle of mass m is confined to an infinitely deep square-well potential:

V(x) = ∞, x ≤ 0, x ≥ a
V(x) = 0, 0  x  a

The normalized eigenfunction, labeled by the quantum number n, are



For any state n, the expectation value of the momentum of the particle is

A. 0

B. 

C. 

D. 

E. 

(GR9277 #51)

Solution:

Infinitely deep square-well potential
→ there is zero probability for particle to be outside the well
→ 〈〉= 0

If 〈〉 ≠ 0 the particle would tend to go to the right or left and leave the well, which is impossible for infinitely deep square-well potential.

Answer: A


Alternative Answer #1:

Eigen function,  ψ_n = λψ_n
The eigenvalue, λ is associated with expectation value:〈Â〉= 〈ψ | Â | ψ〉

Since  is imaginer and ψ_n is real →  the eigenvalue, λ is imaginer = not real = not observable
→ 〈〉= 0


Alternative Answer #2:

since sine and cosine are orthogonal the whole period.

Quantum Mechanics - Infinite Potential Well

See Problem 51

The eigenfunctions satisfy the condition

₀∫^aψ_n*(x) ψ_l (x) dx = δ_nl

δ_nl = 1 if n = l, otherwise δ_nl = 0. This is a statement that the eigenfunctions are

A. Solutions to the Schrodinger equation 
B. Orthonormal
C. Bounded 
D. Linearly dependent
E. Symmetric

(GR9277 #52)

Solution:

Orthonormality = orthogonal and normal

₀∫^aψ_n*(x) ψ_l (x) dx = 〈ψ_n|ψ_l〉= δ_nl

Orthogonal,  δ_nl = 0
Normal, δ_nl = 1 if n = l

Answer: B